Walabuma deposits Br 550 a year for 10 years into his account that pays 8% compounded annually. After 10 years, Walabuma transfers the money into another account that pays 10% compounded quarterly. The money is left in the second account for 8 years.

Answer 1

To solve this problem, we need to calculate the future value of the initial deposit in the first account after 10 years, and then calculate the future value of that amount after another 8 years in the second account.

For the first account, we'll use the formula for compound interest:
A = P * (1 + r/n)^(nt)
where A is the future value, P is the principal amount, r is the annual interest rate in decimal form, n is the number of times the interest is compounded per year, and t is the number of years.

Using the given values:
P = 550 (the annual deposit)
r = 8% = 0.08 (the annual interest rate in decimal form)
n = 1 (compounded annually)
t = 10 (number of years)

Calculating the future value in the first account after 10 years:
A = 550 * (1 + 0.08/1)^(1*10)
A = 550 * (1 + 0.08)^10
A = 550 * (1.08)^10
A = 550 * 2.158927
A ≈ 1187.91

So, after 10 years, the amount in the first account would be approximately Br 1187.91.

Now, we need to calculate the future value of that amount in the second account after 8 years using the same formula:
P = 1187.91 (the amount transferred from the first account)
r = 10% = 0.10 (the annual interest rate in decimal form)
n = 4 (compounded quarterly)
t = 8 (number of years)

Calculating the future value in the second account after 8 years:
A = 1187.91 * (1 + 0.10/4)^(4*8)
A = 1187.91 * (1 + 0.025)^32
A = 1187.91 * (1.025)^32
A ≈ 1187.91 * 2.198218
A ≈ 2613.96

So, after 8 years in the second account, the amount would be approximately Br 2613.96.

Therefore, after 10 years in the first account and another 8 years in the second account, Walabuma would have approximately Br 2613.96.